Copyright © R. R. Dickerson 20101 Lecture 5 AOSC/CHEM 637 Atmospheric Chemistry R. Dickerson...

Copyright © R. R. Dickerson 2010 1

Lecture 5AOSC/CHEM 637

Atmospheric ChemistryR. Dickerson

OUTLINE

KINETICS

Activation Energy

Kinetic Theory of Gases

Calc. Rate Constants w/Collision Theory

Finlayson-Pitts (2000) Ch. 5

Seinfeld and Pandis (2006) Ch. 31


Kinetics continued Activation Energy

The energy hill that reactants must climb in order to produce products; a barrier to thermodynamic equilibrium; for a second order reaction:

ENERGY DIAGRAM

C + DA + B

AB†

ΔH

Ea

A + B → AB†

AB† + M → AB + M†

AB† → C + D

(Transition state or activated complex)

(Sometimes there can be quenching)

(Reaction)

2


ACTIVATION ENERGY

Remember the Van't Hoff (or Gibbs-Helmholtz) equation.

dlnKeq/dT = H/(RT2)

This suggests:

dlnk/dT = Ea /(RT2)

Which is the Arrhenius expression where Ea is the activation energy. If we

integrate both sides:ln(k) = (-Ea/R) 1/T + ln(A)

Where ln(A) is the constant of integration. Rearranging:

k = A e(-Ea/RT)

This is the Arrhenius Equation in which A is the pre-exponential factor, also called the Arrhenius factor, and exp(-Ea/RT) is the Boltzmann factor.

3


KINETIC THEORY OF GASES

When molecules in the gas phase collide they sometimes rearrange their chemical bonds to form new molecules. The rate of formation of the new molecules is determined by the fraction of molecules with sufficient energy to overcome the activation energy barrier.

POSTULATES OF CHEMICAL KINETICS

1. Pressure is the result of molecular collisions.2. Collisions are elastic, i.e. no change in kinetic energy.3. Volume of the molecules << volume occupied by gas.4. Kinetic energy proportional to T and independent of gas, i.e., the same for all gases.

4


Boltzmann Distribution

N1/N0 = e{-(E1 - E0 )/kT}

(Also called Maxwell distribution for ideal gases)

WHERE: N1 = number of particles (molecules) with energy E1

N0 = number of particles (molecules) with energy E0

M = molecular weight

dN/N0 = M/kT exp [{ -Mc-2 }/2kT] c dc

WHERE:c2 = V2 + U2 + W2

SEE: Lavenda, "Brownian Motion," Sci. Amer., 252(2), 70-85, 1985.

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CALCULATING RATE CONSTANTS FROM COLLISION THEORY

From thermodynamics and Arrhenius:

k = A exp(-Ea/RT)

A is a function of diameter, temperature, and mass; its maximum possible value is the frequency of collisions.

WHERE:k = Boltzmann const. = 1.38x10-16 erg/KT = Abs. Temp. (K) d = Diameter of molecules.= reduced mass = {M1 x M2} / {M1 + M2}

A has units of (molecules cm-3)-1 s-1 or cm3 s-1

FrequencyCollision 8kT

A 2

2/1

d

7


Example: Collisions Between Nitrogen Molecules

dN2 = 3.2x10-8 cm

MN2 = 28/6.023x1023 g

For N2 + N2

This is a good estimate for the maximum rate constant for any reaction. Note A is proportional to d2, , T1/2. One would expect the Arrhenius factor to have a T1/2 factor, but this is usually swamped out by the exponential temperature dependence of activation energy. (Remember ergs are g cm2 s-2).

31-10-

28

2/1

23

-16

cm s 2x10

Freq.Collision )102.3(103.2

2981.38x108 A

xx

8


Example calculation of a rate constant

O3 + NO ↔ NO2 + O2

ENERGY DIAGRAM ↑

NO2 + O2O3 + NO

O=O-O ∙ ∙ ∙ N=O

ΔH

Ea

O=O-O ∙ ∙ ∙ N=O is the activate complex

Ea is the activation energy, unknown.

ΔH is the enthalpy of the reaction, known from thermodynamics.

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Rate const for the forward reaction:kf = A exp(-Ea/RT)

We need the enthalpy:

∆Horxn = ∆H f

o prod. - ∆Hf o react.

∆H f = {8.1 + 0.0 - 34.0 - 21.6} = -47.5 kcal/mole

∆Hr = - ∆Hf = + 47.5 kcal/mole

We need the pre-exponential factors Af and Ar

DIAMETERSd(NO) = 0.40 nmd(O3) = 0.46 nm

d(NO2) = 0.46 nm

d(O2) = 0.296 nm

RT

H-E-

rr

a

e A k

:reaction reverse for theconst Rate

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REDUCED MASSES f = 18.5/6.023x1023 g

r = 18.9/6.023x1023 g

Af < Forward collision rate = 3.4x10-10 s-1 cm3

Ar < Reverse collision rate = 2.6x10-10 s-1 cm3

Now we need an estimate of activation energy, Ea

Ear 47.5 kcal/mole

Eaf We know nothing!

This is very slow!

STUDENTS: Calculate the lifetime of NO2 with respect to

conversion to NO at the typical oxygen content of the atmosphere.

1-345

10-r

s cm 1035.6

29898.1

47500- exp 2.6x10 k

:reaction reverse for theconst Rate

x

x

11


To get at kf lets look to thermodynamics.

kf/k r = Keq = exp(-G/RT)

Go = 0 + 12.4 - 21.0 - 39.1 = -47.7 kcal/mole

The products are heavily favored. kf = Keq x kr

But we knew that much from the collision rate already.

The measured rate constant for this reaction is:

34eq 107.5

47700 exp K x

RT

cal/mole 2750E

scm1.9x10 (298) k

scm1500-

exp3.0x10 k

a

1314-f

1312-f

T

13-10f

4534req

scm 106.3k

10x35.6 105.7xk K

x

xx

12


The measured "A" is 113 times smaller than the maximum "A". Why? Not every collision with sufficient energy results in a reaction. The molecules must have the proper orientation.

STERIC FACTOR: A(collisional)/A(actual) = 113

Only one collision in 113 has the proper orientation. Now lets try to calculate a better value for kr. Assume same steric

factor.

Ear = Eaf + Hr = 3000 + 47500

kr = 3.0x10-12 exp(-25500/T) cm3 s-1

= 2.0x10-49 cm3 s-1 at 298 K

Thermodynamics says:

kr = kf/Keq

= 3.2x10-49 cm3 s-1

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The agreement is not too bad, less than a factor of two difference! The thermodynamic value is more likely correct. We cannot measure the reverse rate constant because it is too slow. For example if we took a 1 atm mixture of 50% NO2 and 50% O2 at equilibrium (square brackets represent partial pressure)

the ozone and nitric oxide concentrations would be much too small to measure.

ppt 10 atm2.1x10 ][O[NO]

107.5[0.5][0.5]

]][O[NO K

107.5][NO][O

]][O[NO K

6-18-3

3422eq

34

3

22eq

x

x

Pretty small, so we can say that all the NO and O3 react away.

14


-216KJ/mole

+286 KJ/mole

+428 KJ/mole

Stedman’s Kinetic Kidney

Enthalpy ↑

O2 → 2O H = +498 KJ/mole

k = A e(-Ea/RT)


Lecture 5 Summary

• For simple bimolecular (second order) reactions, the Arrhenius expression describes the temperature dependence of the rate constants.

• Collision theory gives some insight into the limits of rate constants.

• The enthalpy of a reaction provides insight into the activation energy. Molecular oxygen, O2, is not generally an important oxidant in atmospheric reactions.

• The radicals OH and HO2 are the “hit men” of atmospheric chemistry.

Copyright © R. R. Dickerson 20101 Lecture 5 AOSC/CHEM 637 Atmospheric Chemistry R. Dickerson...

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